If y = mx + 4 Is a Tangent to Both the Parabolas y² = 4x and x² = 2by, Then b Is Equal To

JEE Main 2020 — 7 January, Shift 1. Previous Year Question.

Problem

If $y=mx+4$ is a tangent to both the parabolas $y^2=4x$ and $x^2=2by$, then $b$ is equal to:

Key insight. The first parabola has a known tangent formula that immediately pins down $m$; once $m$ is known, tangency to the second parabola becomes an ordinary “discriminant equals zero” condition on $b$.

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Approach

Use the standard tangent condition for $y^2=4ax$, which has a known slope-form tangent $y=mx+\frac{a}m$, to find $m$ by comparing with the given line. Then substitute the now fully-known line into the second parabola’s equation, $x^2=2by$, and use the fact that a tangent line meets a parabola in exactly one point — forcing the resulting quadratic’s discriminant to be zero — to solve for $b$.

Solution

Step 1 — Find m from the first parabola

Writing $y^2=4x$ as $y^2=4ax$ gives $a=1$. Any tangent to this parabola in slope form is $y=mx+\dfrac{a}m=mx+\dfrac1m$. Comparing with the given line $y=mx+4$:

$$\frac1m = 4 \implies m=\frac14$$

Step 2 — Write the fully-determined tangent line

$$y = \frac14x+4$$

Step 3 — Substitute into the second parabola

Substituting this $y$ into $x^2=2by$:

$$x^2 = 2b\left(\frac14x+4\right) = \frac{b}2x+8b$$

Rearranging so everything is on one side:

$$x^2-\frac{b}2x-8b = 0$$

Multiplying through by 2 to clear the fraction:

$$2x^2-bx-16b = 0$$

Step 4 — Apply the tangency condition

A tangent line touches a parabola at exactly one point, so this quadratic in $x$ must have equal roots — its discriminant must be zero. For $2x^2-bx-16b=0$, with $A=2$, $B=-b$, $C=-16b$:

$$B^2-4AC = 0 \implies (-b)^2-4(2)(-16b) = 0$$

$$b^2+128b = 0$$

Step 5 — Solve for b

$$b(b+128) = 0 \implies b=0 \text{ or } b=-128$$

The value $b=0$ must be rejected: substituting $b=0$ into $x^2=2by$ collapses it to $x^2=0$, which isn’t a parabola at all, just a repeated line. So the only valid value is

$$b=-128$$

Answer

Common mistakes

• Forgetting to reject b=0. The discriminant equation always has $b=0$ as one root here, since it makes the constant term of the quadratic vanish too — but this value degenerates the second “parabola” entirely, so it isn’t a genuine solution.
• Sign errors when multiplying through by 2. Going from $x^2-\frac{b}2x-8b=0$ to $2x^2-bx-16b=0$ is a common place to drop a factor of 2 on one term but not another.
• Mixing up which parabola’s tangent formula to use first. The slope-form tangent $y=mx+\frac{a}m$ only applies directly to $y^2=4ax$; for $x^2=2by$, the correct move is to substitute the already-known line and apply the discriminant condition, not to look for an analogous ready-made formula.

Practise next

• If $y=mx+6$ is a tangent to both $y^2=8x$ and $x^2=2by$, find $b$ using the same two-step approach.